Integral Calculus: Basic Integration of cotxln(sinx)

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Homework Statement


evaluate the integral.

Homework Equations


[itex]\int (cotx)[ln(sinx)] dx[/itex]

how would i do this using basic integration techniques from calculus 1? i am not allowed to use, int by parts, partial fractions, trig sub, etc..

The Attempt at a Solution


i tried doing a u=sinx but don't know how to get the cotx.
 
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whatlifeforme said:

Homework Statement


evaluate the integral.

Homework Equations


[itex]\int (cotx)[ln(sinx)][/itex]

how would i do this using basic integration techniques from calculus 1? i am not allowed to use, int by parts, partial fractions, trig sub, etc..

The Attempt at a Solution


i tried doing a u=sinx but don't know how to get the cotx.
Where's the "dx" ?

[itex]\displaystyle \int \cot(x)\ln(\sin(x))\,dx[/itex]

This may help. [itex]\displaystyle \ \cot(x)=\frac{\cos(x)}{\sin(x)}[/itex]
 
thanks.

so i have:

u=sinx; du=cosx

[itex]\int \frac{ln(u)}{u} du[/itex]

v=lnu; dv=(1/u) du

[itex]\int v dv[/itex] = [itex]\frac{v^2}{2}[/itex] = [itex]\frac{(lnu)^2}{2}[/itex]

= [itex]\frac{[ln(sinx)]^2}{2}[/itex]
 
whatlifeforme said:
thanks.

so i have:

u=sinx; du=cosx

[itex]\int \frac{ln(u)}{u} du[/itex]

v=lnu; dv=(1/u) du

[itex]\int v dv[/itex] = [itex]\frac{v^2}{2}[/itex] = [itex]\frac{(lnu)^2}{2}[/itex]

= [itex]\frac{[ln(sinx)]^2}{2}[/itex]

Looks good to me, other than the +C.